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Math Help - Integer part

  1. #1
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    Integer part

    Hi everyone.

    Show that \forall{x\in{\mathbb{Z^+}\cup{0}}} then [\frac{n}{2}]+[-\frac{n}{2}]=n

    Thanks so much.
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  2. #2
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    Re: Integer part

    You mean \forall n\in\mathbb Z^+\cup\{0\}, \left\lfloor\frac n2\right\rfloor+\left|\left\lfloor-\frac n2\right\rfloor\right|=n?

    If n is even, say n=2k, \left\lfloor\frac n2\right\rfloor+\left|\left\lfloor-\frac n2\right\rfloor\right|=k+\left|-k\right|=2k=n.

    If n is odd, say n=2k+1, \left\lfloor\frac n2\right\rfloor+\left|\left\lfloor-\frac n2\right\rfloor\right|=\lfloor k+\tfrac12\rfloor+\left|\lfloor-k-\tfrac12\rfloor\right|=k+|-k-1|=2k+1=n.
    Last edited by Sylvia104; April 6th 2012 at 11:27 PM.
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  3. #3
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    Re: Integer part

    Quote Originally Posted by Sylvia104 View Post
    You mean \forall n\in\mathbb Z^+\cup\{0\}
    Thanks, you're right.

    But I don't know if you use absolute value.
    Last edited by Fernando; April 7th 2012 at 12:57 PM.
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  4. #4
    Member Sylvia104's Avatar
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    Re: Integer part

    Without absolute value, \left\lfloor\frac n2\right\rfloor+\left\lfloor-\frac n2\right\rfloor=0 if n is even and -1 if n is odd.
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  5. #5
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    Re: Integer part

    Thanks, maybe the exercise has a little mistake.
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  6. #6
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    Re: Integer part

    Quote Originally Posted by Fernando View Post
    Show that \forall{x\in{\mathbb{Z^+}\cup{0}}} then [\frac{n}{2}]+[-\frac{n}{2}]=n
    As pointed out that is false.
    However \forall{n\in{\mathbb{Z^+}\cup{0}}} then [\frac{n}{2}]-[-\frac{n}{2}]=n is true.
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  7. #7
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    Re: Integer part

    Quote Originally Posted by Plato View Post
    However \forall{n\in{\mathbb{Z^+}\cup{0}}} then [\frac{n}{2}]-[-\frac{n}{2}]=n is true.
    Yes, You have the reason!

    The proof is similar as it was done by Silvia104 , right?
    Remark: I'm sorry for me grammar.
    Thanks!
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